Cash and Carry

A favorite problem of Lewis Carroll involves a customer trying to complete a purchase using pre-decimal currency. He wants to buy 7s. 3d. worth of goods, but he has only a half-sovereign (10s.), a florin (2s.), and a sixpence. The shopkeeper can’t give him change, as he himself has only a crown (5s.), a shilling, and a penny. As they’re puzzling over this a friend enters the shop with a double-florin (4s.), a half-crown (2s. 6d.), a fourpenny-bit, and a threepenny-bit. Can the three of them negotiate the transaction?

Happily, they can. They pool their money on the counter, and the shopkeeper takes the half-sovereign, the sixpence, the half-crown, and the fourpenny-bit; the customer takes the double-florin, the shilling, and threepenny-bit as change; and the friend takes the florin, the crown, and the penny.

“There are other combinations,” writes John Fisher in The Magic of Lewis Carroll, “but this is the most logistically pleasing, as it will be seen that not one of the three persons retains any one of his own coins.”

Related: From Henry Dudeney, a magic square:

(Strand, December 1896)

September 6, 2010November 4, 2010 | Puzzles · Science & Math

Counter Play

A devilish puzzle by Lee Sallows:

lee sallows counter play

In the diagram above, nine numbered counters occupy the cells of a 3×3 checkerboard so as to form a magic square. Any 3 counters lying in a straight line add up to 15. There are 8 of these collinear triads.

Reposition the counters (again, one to each cell) to yield 8 new collinear triads, but now showing a common sum of 16 rather than 15.

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September 4, 2010January 25, 2013 | Puzzles

Husbands and Wives

This problem dates from at least 1774; this version appeared in the American Mathematical Monthly of December 1902:

Three Dutchmen and their wives went to market to buy hogs. The names of the men were Hans, Klaus, and Hendricks, and of the women, Gertrude, Anna, and Katrine; but it was not known which was the wife of each man. They each bought as many hogs as each man or woman paid shillings for each hog, and each man spent three guineas more than his wife. Hendricks bought 23 hogs more than Gertrude, and Klaus bought 11 more than Katrine. What was the name of each man’s wife?

(There are 21 shillings in a guinea.)

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The American Mathematical Monthly‘s published solution to Husbands and Wives: Let x represent the number of one of the women’s hogs, and y the number of her husband’s; then by the conditions of the problem y² = x² + 63. Consequently x² + 63 must be an integer, since represents the number of hogs. The equation y² – x² = 63 or (y + x) (y – x) = 63 admits of three solutions, viz., 63 × 1, 21 × 3, and 9 × 7: whence y = 32, x = 31; y = 12, x = 9; y = 8, x = 1. Since 32 = 9 + 23 and 12 = 1 + 11, 32 belongs to Hendricks, 12 to Klaus, 9 belongs to Katrine, 1 to Gertrude. Therefore 32 Hendricks, 31 Anna; 12 Klaus, 9 Katrine; 8 Hans, 1 Gertrude. Hence, Hendricks is Anna’s husband, Klaus is Katrine’s husband, and Hans is Gertrude’s husband. The numbers here are correct, but the names are confused. Correct are: Hendricks 32 Gertrude 9 Klaus 12 Katrine 1 Hans 8 Anna 31 So Hendricks is married to Anna, Klaus to Gertrude, and Hans to Katrine. (Thanks, Shrey.)

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September 2, 2010January 25, 2013 | Puzzles

Tight Squeeze

By T.R. Dowson. White to mate in 21 moves:

tight squeeze chess puzzle

It’s not as hard as it sounds, though it’s a bit like a square dance in a submarine.

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1. Ng3 Kb3 2. Rh1 Ka3 3. Ng1 Kb3 4. Rf3 Ka3 5. Bf2 Kb3 6. Re3 Ka3 7. Nf3 Kb3 8. Rg1 Ka3 9. Nh1 Kb3 10. Bg3 Ka3 11. Nf2 Kb3 12. Rh1 Ka3 13. Ng1 Kb3 14. Rf3 Ka3 15. Ke3 Kb3 16. Nd3 Ka3 17. Kf2 Kb3 18. Re3 Ka3 19. Kf3 Kb3 20. Nf2 Ka3 21. Rxc3# (Thanks to Matt for this solution.)

August 31, 2010January 25, 2013 | Puzzles

Reading Matter

In a certain library, no two books contain the same number of words, and the total number of books is greater than number of words in the largest book.

How many words does one of the books contain, and what is it about?

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At least one of the books must be entirely blank, about nothing. Or, as someone just wrote to me, “It’s about things that men know about women.” From Robert Morrison Abraham, Diversions and Pastimes, 1935.

August 27, 2010January 25, 2013 | Puzzles

Current Affairs

At 1:17 one afternoon a canoeist left his riverside camp and paddled upstream at 4 mph against a current of 1.5 mph. At 2:05 he passed a corked bottle floating downstream and noticed that it contained a message. He paddled some distance further but finally couldn’t help himself — he turned around and paddled after the bottle. He caught it just as it reached his camp. The message read:

HOW FAR DID YOU GET FROM CAMP BEFORE YOU GAVE IN TO YOUR CURIOSITY?

“There is no reason why the camper should have paid any attention to this odd message, but you know how these things are,” writes Geoffrey Mott-Smith in Mathematical Puzzles for Beginners and Enthusiasts (1946). The camper had noticed a landmark at the point upstream where he’d turned around, so he was able to measure the distance the next day. But he could have reasoned the thing out from the facts. Can you?

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The camper reached a point 3 2/3 miles from his camp. First he paddled upstream at a net rate of 2.5 mph for 48 minutes and thus met the bottle at a point 2 miles above camp. From that point the bottle took 80 minutes to float down to camp. The canoeist spent half that time paddling upstream and half downstream (the two times must be equal). In 40 minutes paddling upstream he traveled another 1 2/3 miles, for a total of 3 2/3 miles. A simple way to see why the 80 minutes must be halved is to imagine that we can turn off the river’s current. In that case the canoeist is simply paddling at a steady rate away from the bottle and back to it.

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August 24, 2010January 25, 2013 | Puzzles

Clutch Cargo

Prove that the number of people who shake hands an odd number of times at the opera next Thursday will be even.

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Call a person an Odd if he has shaken hands an odd number of times (at any given moment). An Even has shaken hands an even number of times. At the start, the number of Odds is zero, an even number. As the evening progresses, three types of handshake take place: An Odd shakes with an Odd. In this case they both become Evens; the total number of Odds decreases by 2. An Even shakes with an Even. Both become Odds, and the number of Odds increases by 2. An Even shakes with an Odd. The Even becomes an Odd, and the Odd becomes an Even, so the total number of Odds remains unchanged. Because the number of Odds can change only by an even number, the number of people who have shaken hands an odd number of times remains even — not only at the end of the evening, but at every moment throughout! A Norwegian reader adds: “It might possibly be of some minor interest that in Norway, Odd and Even are both widespread men’s names. Fittingly, as of the most recently updated name statistics (2009), there are 22821 Norwegian men with Odd as their first first name, and 5846 with Even as their first first name. (And 29 are called Odd Even.)”

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August 21, 2010January 25, 2013 | Puzzles

Black and White

By W. Atkinson. White to mate in two moves.

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1. Qe4 and White wins by 2. Qxe5#, 2. c5#, 2. Qg6#, or 2. Qxd5#. From Charles F. Stubbs, Canadian Chess Problems, 1890.

August 18, 2010January 25, 2013 | Puzzles

The Cigar Puzzle

From Henry Dudeney:

Two men are seated at a square-topped table. One places an ordinary cigar (flat at one end, pointed at the other) on the table, then the other does the same, and so on alternately, a condition being that no cigar shall touch another. Which player should succeed in placing the last cigar, assuming that they each will play in the best possible manner? The size of the table top and the size of the cigar are not given, but in order to exclude the ridiculous answer that the table might be so diminutive as only to take one cigar, we will say that the table must not be less than 2 feet square and the cigar not more than 4-1/2 inches long. With those restrictions you may take any dimensions you like. Of course we assume that all the cigars are exactly alike in every respect. Should the first player, or the second player, win?

Geoffrey Mott-Smith writes, “I cannot resist narrating that I first became acquainted with this gem while reading in bed, and that like an illustrious precursor I startled the household by jumping out of bed, dancing about crying ‘Eureka! Eureka!'”

What had he seen?

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The first player must place his first cigar on end in the exact centre of the table, as indicated by the little circle. Now, whatever the second player may do throughout, the first player must always repeat it in an exactly diametrically opposite position. Thus, if the second player places a cigar at A, I put one at AA; he places one at B, I put one at BB; he places one at C, I put one at CC; he places one at D, I put one at DD; he places one at E, I put one at EE; and so on until no more cigars can be placed without touching. “I smoke ten to fifteen cigars a day,” said George Burns. “At my age I have to hold on to something.”

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August 13, 2010January 25, 2013 | Puzzles

The Potato Paradox

You have 100 pounds of Martian potatoes, which are 99 percent water by weight. You let them dehydrate until they’re 98 percent water. How much do they weigh now?

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Perhaps surprisingly, the potatoes now weigh 50 pounds. The sack of potatoes originally weighed 100 pounds, and 1 percent of its weight, or 1 pound, was non-water. If that 1 pound now makes up 2 percent of the total weight, then the total weight must be 50 pounds. Not a paradox, exactly, but many people find it counterintuitive.

August 10, 2010May 9, 2013 | Puzzles